我需要在程序中更改哪些内容才能计算素数的上限？
我正在CUDA编写自己的Eratosthenes筛子实施方案，而且我已经碰壁了。 到目前为止，该算法似乎适用于少数（低于8500万）。
然而，当我尝试计算高达1亿，20亿，30亿的素数时，系统冻结（当它在CUDA设备中计算时），然后几秒钟后，我的linux机器恢复正常（解冻），但CUDA程序崩溃时出现以下错误消息：
CUDA error at prime.cu:129 code=6(cudaErrorLaunchTimeout) "cudaDeviceSynchronize()"
我有一个GTX 780（3 GB）并且我在一个char数组中分配了sieves，所以如果我要计算高达100,000的素数，它将在设备中分配100,000个字节。
我认为GPU可以容纳多达30亿个数字，因为它有3 GB的内存，但是，它只允许我做8500万个顶部（8500万字节= 0.08 GB）
这是我的 prime.cu 代码：
#include <stdio.h>
#include <helper_cuda.h> // checkCudaErrors() - NVIDIA_CUDA-6.0_Samples/common/inc
// #include <cuda.h>
// #include <cuda_runtime_api.h>
// #include <cuda_runtime.h>
typedef unsigned long long int uint64_t;
/******************************************************************************
* kernel that initializes the 1st couple of values in the primes array.
******************************************************************************/
__global__ static void sieveInitCUDA(char* primes)
{
primes[0] = 1; // value of 1 means the number is NOT prime
primes[1] = 1; // numbers "0" and "1" are not prime numbers
}
/******************************************************************************
* kernel for sieving the even numbers starting at 4.
******************************************************************************/
__global__ static void sieveEvenNumbersCUDA(char* primes, uint64_t max)
{
uint64_t index = blockIdx.x * blockDim.x + threadIdx.x + threadIdx.x + 4;
if (index < max)
primes[index] = 1;
}
/******************************************************************************
* kernel for finding prime numbers using the sieve of eratosthenes
* - primes: an array of bools. initially all numbers are set to "0".
* A "0" value means that the number at that index is prime.
* - max: the max size of the primes array
* - maxRoot: the sqrt of max (the other input). we don't wanna make all threads
* compute this over and over again, so it's being passed in
******************************************************************************/
__global__ static void sieveOfEratosthenesCUDA(char *primes, uint64_t max,
const uint64_t maxRoot)
{
// get the starting index, sieve only odds starting at 3
// 3,5,7,9,11,13...
/* int index = blockIdx.x * blockDim.x + threadIdx.x + threadIdx.x + 3; */
// apparently the following indexing usage is faster than the one above. Hmm
int index = blockIdx.x * blockDim.x + threadIdx.x + 3;
// make sure index won't go out of bounds, also don't start the execution
// on numbers that are already composite
if (index < maxRoot && primes[index] == 0)
{
// mark off the composite numbers
for (int j = index * index; j < max; j += index)
{
primes[j] = 1;
}
}
}
/******************************************************************************
* checkDevice()
******************************************************************************/
__host__ int checkDevice()
{
// query the Device and decide on the block size
int devID = 0; // the default device ID
cudaError_t error;
cudaDeviceProp deviceProp;
error = cudaGetDevice(&devID);
if (error != cudaSuccess)
{
printf("CUDA Device not ready or not supported\n");
printf("%s: cudaGetDevice returned error code %d, line(%d)\n", __FILE__, error, __LINE__);
exit(EXIT_FAILURE);
}
error = cudaGetDeviceProperties(&deviceProp, devID);
if (deviceProp.computeMode == cudaComputeModeProhibited || error != cudaSuccess)
{
printf("CUDA device ComputeMode is prohibited or failed to getDeviceProperties\n");
return EXIT_FAILURE;
}
// Use a larger block size for Fermi and above (see compute capability)
return (deviceProp.major < 2) ? 16 : 32;
}
/******************************************************************************
* genPrimesOnDevice
* - inputs: limit - the largest prime that should be computed
* primes - an array of size [limit], initialized to 0
******************************************************************************/
__host__ void genPrimesOnDevice(char* primes, uint64_t max)
{
int blockSize = checkDevice();
if (blockSize == EXIT_FAILURE)
return;
char* d_Primes = NULL;
int sizePrimes = sizeof(char) * max;
uint64_t maxRoot = sqrt(max);
// allocate the primes on the device and set them to 0
checkCudaErrors(cudaMalloc(&d_Primes, sizePrimes));
checkCudaErrors(cudaMemset(d_Primes, 0, sizePrimes));
// make sure that there are no errors...
checkCudaErrors(cudaPeekAtLastError());
// setup the execution configuration
dim3 dimBlock(blockSize);
dim3 dimGrid((maxRoot + dimBlock.x) / dimBlock.x);
dim3 dimGridEvens(((max + dimBlock.x) / dimBlock.x) / 2);
//////// debug
#ifdef DEBUG
printf("dimBlock(%d, %d, %d)\n", dimBlock.x, dimBlock.y, dimBlock.z);
printf("dimGrid(%d, %d, %d)\n", dimGrid.x, dimGrid.y, dimGrid.z);
printf("dimGridEvens(%d, %d, %d)\n", dimGridEvens.x, dimGridEvens.y, dimGridEvens.z);
#endif
// call the kernel
// NOTE: no need to synchronize after each kernel
// http://stackoverflow.com/a/11889641/2261947
sieveInitCUDA<<<1, 1>>>(d_Primes); // launch a single thread to initialize
sieveEvenNumbersCUDA<<<dimGridEvens, dimBlock>>>(d_Primes, max);
sieveOfEratosthenesCUDA<<<dimGrid, dimBlock>>>(d_Primes, max, maxRoot);
// check for kernel errors
checkCudaErrors(cudaPeekAtLastError());
checkCudaErrors(cudaDeviceSynchronize());
// copy the results back
checkCudaErrors(cudaMemcpy(primes, d_Primes, sizePrimes, cudaMemcpyDeviceToHost));
// no memory leaks
checkCudaErrors(cudaFree(d_Primes));
}
测试此代码：
int main()
{
int max = 85000000; // 85 million
char* primes = malloc(max);
// check that it allocated correctly...
memset(primes, 0, max);
genPrimesOnDevice(primes, max);
// if you wish to display results:
for (uint64_t i = 0; i < size; i++)
{
if (primes[i] == 0) // if the value is '0', then the number is prime
{
std::cout << i; // use printf if you are using c
if ((i + 1) != size)
std::cout << ", ";
}
}
free(primes);
}
答案 0 :(得分：2)
此错误：
CUDA error at prime.cu:129 code=6(cudaErrorLaunchTimeout) "cudaDeviceSynchronize()"
并不一定意味着什么。它不一定是数字限制或计算错误，而是系统对内核允许运行的时间量施加的限制。 Linux和Windows都可以有这样的看门狗定时器。
如果您想在linux案例中解决此问题，请查看this document。
你没有提到它，但我认为你的GTX780也是一个（这个）显示器。在这种情况下，默认情况下内核有一个时间限制。如果您可以使用其他设备作为显示器，则重新配置您的设备，让X不使用GTX780，如链接中所述。如果您没有其他GPU用于显示，那么唯一的选择是修改链接文档中指示的交互设置，如果您想运行长时间运行的内核。在这种情况下，键盘/鼠标/显示器将在内核运行时变得无响应。如果您的内核运行时间太长，则可能难以恢复计算机，并且可能需要重新启动硬盘。 （您也可以通过SSH连接到计算机，并终止使用GPU进行CUDA的过程。）